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ISEE Middle Level Quantitative : ISEE Middle Level (grades 7-8) Quantitative Reasoning

Study concepts, example questions & explanations for ISEE Middle Level Quantitative

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All ISEE Middle Level Quantitative Resources

6 Diagnostic Tests 110 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #24 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Solve:

$\begin{array}{r}\overset{~ \space{\ }}{\ 8} \space{\,}5\\ \times\phantom{0}8\space{\,}8 \\ \hline \phantom{\,}\end{array}$

Possible Answers:

7470

74800

7480

7490

Correct answer:

7480

Explanation:

In order to solve this problem using multiplication, we must multiply each digit of the multiplier by each digit of the multiplicand to get the answer, which is called the product.

$\frac{\begin{array}[b]{r}\textup{multiplicand}\\ \times\ \ \ \ \textup{multiplier}\end{array}}{\ \ \ \ \textup{product}}$

For this problem, 88 is the multiplier and 85 is the multiplicand. We start with the multiplier, and multiply the digit in the ones place by each digit of the multiplicand.

First, we multiply 8 and 5

$\begin{array}{r}\overset{4 \space{\ }}{\ 8} \space{\,}5\\ \times \phantom{0} 8\space{\,}8\\ \hline \phantom{\,} \,0\end{array}$

Then, we multiply 8 and 8 and add the 4 that was carried

$\begin{array}{r}\overset{4 \space{\ }}{\ 8} \space{\,}5\\ \times \phantom{0} 8\space{\,}8 \\ \hline \phantom{\,}6\,8\,0\end{array}$

Now, we multiply the digit in the tens place of the multiplier by each digit of the multiplicand. In order to do this, we need to start a new line in our math problem and put a 0 as a place holder in the ones position.

$\begin{array}{r}\overset{~ \space{\ }}{\ 8} \space{\,}5\\ \times \phantom{0} 8\space{\,}8 \\ \hline \phantom{\,}6\,8\,0 \\ \, 0 \end{array}$

Next, we multiply 8 and 5

$\begin{array}{r}\overset{4 \space{\ }}{\ 8} \space{\,}5\\ \times \phantom{0} 8\space{\,}8 \\ \hline \phantom{\,}6\,8\,0 \\ \, 0 \, 0\end{array}$

Then, we multiply 8 and 8and add the 4 that was carried

$\begin{array}{r}\overset{4 \space{\ }}{\ 8} \space{\,}5\\ \times \phantom{0} 8\space{\,}8 \\ \hline \phantom{\,}6\,8\,0 \\ \, 6 \, 8 \, 0 \, 0\end{array}$

Finally, we add the two products together to find our final answer

$\begin{array}{r}\overset{4 \space{\ }}{\ 8} \space{\,}5\\ \times \phantom{0} 8\space{\,}8 \\ \hline \phantom{\,}6\,8\,0 \\+\, 6 \, 8 \, 0 \, 0\\ \hline 7\, 4 \, 8\, 0\end{array}$

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Example Question #25 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Solve:

$\begin{array}{r}\overset{~ \space{\ }}{\ 8} \space{\,}7\\ \times\phantom{0}3\space{\,}3 \\ \hline \phantom{\,}\end{array}$

Possible Answers:

28710

2861

2871

2881

Correct answer:

2871

Explanation:

In order to solve this problem using multiplication, we must multiply each digit of the multiplier by each digit of the multiplicand to get the answer, which is called the product.

$\frac{\begin{array}[b]{r}\textup{multiplicand}\\ \times\ \ \ \ \textup{multiplier}\end{array}}{\ \ \ \ \textup{product}}$

For this problem, 33 is the multiplier and 87 is the multiplicand. We start with the multiplier, and multiply the digit in the ones place by each digit of the multiplicand.

First, we multiply 3 and 7

$\begin{array}{r}\overset{2 \space{\ }}{\ 8} \space{\,}7\\ \times \phantom{0} 3\space{\,}3\\ \hline \phantom{\,} \,1\end{array}$

Then, we multiply 3 and 8 and add the 2 that was carried

$\begin{array}{r}\overset{2 \space{\ }}{\ 8} \space{\,}7\\ \times \phantom{0} 3\space{\,}3 \\ \hline \phantom{\,}2\,6\,1\end{array}$

$\begin{array}{r}\overset{~ \space{\ }}{\ 8} \space{\,}7\\ \times \phantom{0} 3\space{\,}3 \\ \hline \phantom{\,}2\,6\,1 \\ \, 0 \end{array}$

Next, we multiply 3 and 7

$\begin{array}{r}\overset{2 \space{\ }}{\ 8} \space{\,}7\\ \times \phantom{0} 3\space{\,}3 \\ \hline \phantom{\,}2\,6\,1 \\ \, 1 \, 0\end{array}$

Then, we multiply 3 and 8and add the 2 that was carried

$\begin{array}{r}\overset{2 \space{\ }}{\ 8} \space{\,}7\\ \times \phantom{0} 3\space{\,}3 \\ \hline \phantom{\,}2\,6\,1 \\ \, 2 \, 6 \, 1 \, 0\end{array}$

Finally, we add the two products together to find our final answer

$\begin{array}{r}\overset{2 \space{\ }}{\ 8} \space{\,}7\\ \times \phantom{0} 3\space{\,}3 \\ \hline \phantom{\,}2\,6\,1 \\+\, 2 \, 6 \, 1 \, 0\\ \hline 2\, 8 \, 7\, 1\end{array}$

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Example Question #26 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Solve:

$\begin{array}{r}\overset{~ \space{\ }}{\ 4} \space{\,}4\\ \times\phantom{0}6\space{\,}3 \\ \hline \phantom{\,}\end{array}$

Possible Answers:

2782

27720

2772

2762

Correct answer:

2772

Explanation:

In order to solve this problem using multiplication, we must multiply each digit of the multiplier by each digit of the multiplicand to get the answer, which is called the product.

$\frac{\begin{array}[b]{r}\textup{multiplicand}\\ \times\ \ \ \ \textup{multiplier}\end{array}}{\ \ \ \ \textup{product}}$

For this problem, 63 is the multiplier and 44 is the multiplicand. We start with the multiplier, and multiply the digit in the ones place by each digit of the multiplicand.

First, we multiply 3 and 4

$\begin{array}{r}\overset{1 \space{\ }}{\ 4} \space{\,}4\\ \times \phantom{0} 6\space{\,}3\\ \hline \phantom{\,} \,2\end{array}$

Then, we multiply 3 and 4 and add the 1 that was carried

$\begin{array}{r}\overset{1 \space{\ }}{\ 4} \space{\,}4\\ \times \phantom{0} 6\space{\,}3 \\ \hline \phantom{\,}1\,3\,2\end{array}$

$\begin{array}{r}\overset{~ \space{\ }}{\ 4} \space{\,}4\\ \times \phantom{0} 6\space{\,}3 \\ \hline \phantom{\,}1\,3\,2 \\ \, 0 \end{array}$

Next, we multiply 6 and 4

$\begin{array}{r}\overset{2 \space{\ }}{\ 4} \space{\,}4\\ \times \phantom{0} 6\space{\,}3 \\ \hline \phantom{\,}1\,3\,2 \\ \, 4 \, 0\end{array}$

Then, we multiply 6 and 4and add the 2 that was carried

$\begin{array}{r}\overset{2 \space{\ }}{\ 4} \space{\,}4\\ \times \phantom{0} 6\space{\,}3 \\ \hline \phantom{\,}1\,3\,2 \\ \, 2 \, 6 \, 4 \, 0\end{array}$

Finally, we add the two products together to find our final answer

$\begin{array}{r}\overset{2 \space{\ }}{\ 4} \space{\,}4\\ \times \phantom{0} 6\space{\,}3 \\ \hline \phantom{\,}1\,3\,2 \\+\, 2 \, 6 \, 4 \, 0\\ \hline 2\, 7 \, 7\, 2\end{array}$

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Example Question #27 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Solve:

$\begin{array}{r}\overset{~ \space{\ }}{\ 8} \space{\,}4\\ \times\phantom{0}5\space{\,}1 \\ \hline \phantom{\,}\end{array}$

Possible Answers:

42840

4274

4294

4284

Correct answer:

4284

Explanation:

In order to solve this problem using multiplication, we must multiply each digit of the multiplier by each digit of the multiplicand to get the answer, which is called the product.

$\frac{\begin{array}[b]{r}\textup{multiplicand}\\ \times\ \ \ \ \textup{multiplier}\end{array}}{\ \ \ \ \textup{product}}$

For this problem, 51 is the multiplier and 84 is the multiplicand. We start with the multiplier, and multiply the digit in the ones place by each digit of the multiplicand.

First, we multiply 1 and 4

$\begin{array}{r}\overset{~ \space{\ }}{\ 8} \space{\,}4\\ \times \phantom{0} 5\space{\,}1 \\ \hline \phantom{\,} \,4\end{array}$

Then, we multiply 1 and 8

$\begin{array}{r}\overset{~ \space{\ }}{\ 8} \space{\,}4\\ \times \phantom{0} 5\space{\,}1 \\ \hline \phantom{\,} \,8\,4\end{array}$

$\begin{array}{r}\overset{~ \space{\ }}{\ 8} \space{\,}4\\ \times \phantom{0} 5\space{\,}1 \\ \hline \phantom{\,} \,8\,4 \\ \, 0 \end{array}$

Next, we multiply 5 and 4

$\begin{array}{r}\overset{2 \space{\ }}{\ 8} \space{\,}4\\ \times \phantom{0} 5\space{\,}1 \\ \hline \phantom{\,} \,8\,4 \\ \, 0 \, 0\end{array}$

Then, we multiply 5 and 8and add the 2 that was carried

$\begin{array}{r}\overset{2 \space{\ }}{\ 8} \space{\,}4\\ \times \phantom{0} 5\space{\,}1 \\ \hline \phantom{\,} \,8\,4 \\ \, 4 \, 2 \, 0 \, 0\end{array}$

Finally, we add the two products together to find our final answer

$\begin{array}{r}\overset{2 \space{\ }}{\ 8} \space{\,}4\\ \times \phantom{0} 5\space{\,}1 \\ \hline \phantom{\,} \,8\,4 \\+\, 4 \, 2 \, 0 \, 0\\ \hline 4\, 2 \, 8\, 4\end{array}$

Report an Error

Example Question #28 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Solve:

$\begin{array}{r}\overset{~ \space{\ }}{\ 4} \space{\,}2\\ \times\phantom{0}2\space{\,}8 \\ \hline \phantom{\,}\end{array}$

Possible Answers:

1176

11760

1186

1166

Correct answer:

1176

Explanation:

In order to solve this problem using multiplication, we must multiply each digit of the multiplier by each digit of the multiplicand to get the answer, which is called the product.

$\frac{\begin{array}[b]{r}\textup{multiplicand}\\ \times\ \ \ \ \textup{multiplier}\end{array}}{\ \ \ \ \textup{product}}$

For this problem, 28 is the multiplier and 42 is the multiplicand. We start with the multiplier, and multiply the digit in the ones place by each digit of the multiplicand.

First, we multiply 8 and 2

$\begin{array}{r}\overset{1 \space{\ }}{\ 4} \space{\,}2\\ \times \phantom{0} 2\space{\,}8\\ \hline \phantom{\,} \,6\end{array}$

Then, we multiply 8 and 4 and add the 1 that was carried

$\begin{array}{r}\overset{1 \space{\ }}{\ 4} \space{\,}2\\ \times \phantom{0} 2\space{\,}8 \\ \hline \phantom{\,}3\,3\,6\end{array}$

$\begin{array}{r}\overset{~ \space{\ }}{\ 4} \space{\,}2\\ \times \phantom{0} 2\space{\,}8 \\ \hline \phantom{\,}3\,3\,6 \\ \, 0 \end{array}$

Next, we multiply 2 and 2

$\begin{array}{r}\overset{~ \space{\ }}{\ 4} \space{\,}2\\ \times \phantom{0} 2\space{\,}8 \\ \hline \phantom{\,}3\,3\,6 \\ \, 4 \, 0\end{array}$

Then, we multiply 2 and 4

$\begin{array}{r}\overset{~ \space{\ }}{\ 4} \space{\,}2\\ \times \phantom{0} 2\space{\,}8 \\ \hline \phantom{\,}3\,3\,6 \\ \, 8 \, 4 \, 0\end{array}$

Finally, we add the two products together to find our final answer

$\begin{array}{r}\overset{~ \space{\ }}{\ 4} \space{\,}2\\ \times \phantom{0} 2\space{\,}8 \\ \hline \phantom{\,}3\,3\,6 \\+\, 8 \, 4 \, 0\\ \hline 1\, 1 \, 7\, 6\end{array}$

Report an Error

Example Question #29 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Solve:

$\begin{array}{r}\overset{~ \space{\ }}{\ 4} \space{\,}5\\ \times\phantom{0}2\space{\,}0 \\ \hline \phantom{\,}\end{array}$

Possible Answers:

890

900

9000

910

Correct answer:

900

Explanation:

In order to solve this problem using multiplication, we must multiply each digit of the multiplier by each digit of the multiplicand to get the answer, which is called the product.

$\frac{\begin{array}[b]{r}\textup{multiplicand}\\ \times\ \ \ \ \textup{multiplier}\end{array}}{\ \ \ \ \textup{product}}$

For this problem, 20 is the multiplier and 45 is the multiplicand. We start with the multiplier, and multiply the digit in the ones place by each digit of the multiplicand.

First, we multiply 0 and 5

$\begin{array}{r}\overset{~ \space{\ }}{\ 4} \space{\,}5\\ \times \phantom{0} 2\space{\,}0 \\ \hline \phantom{\,} \,0\end{array}$

Then, we multiply 0 and 4

$\begin{array}{r}\overset{~ \space{\ }}{\ 4} \space{\,}5\\ \times \phantom{0} 2\space{\,}0 \\ \hline \phantom{\,} \,0\,0\end{array}$

$\begin{array}{r}\overset{~ \space{\ }}{\ 4} \space{\,}5\\ \times \phantom{0} 2\space{\,}0 \\ \hline \phantom{\,} \,0\,0 \\ \, 0 \end{array}$

Next, we multiply 2 and 5

$\begin{array}{r}\overset{1 \space{\ }}{\ 4} \space{\,}5\\ \times \phantom{0} 2\space{\,}0 \\ \hline \phantom{\,} \,0\,0 \\ \, 0 \, 0\end{array}$

Then, we multiply 2 and 4and add the 1 that was carried

$\begin{array}{r}\overset{1 \space{\ }}{\ 4} \space{\,}5\\ \times \phantom{0} 2\space{\,}0 \\ \hline \phantom{\,} \,0\,0 \\ \, 9 \, 0 \, 0\end{array}$

Finally, we add the two products together to find our final answer

$\begin{array}{r}\overset{1 \space{\ }}{\ 4} \space{\,}5\\ \times \phantom{0} 2\space{\,}0 \\ \hline \phantom{\,} \,0\,0 \\+\, 9 \, 0 \, 0\\ \hline 9\, 0 \, 0\end{array}$

Report an Error

Example Question #552 : Number & Operations In Base Ten

Solve:

$\begin{array}{r}\overset{~ \space{\ }}{\ 3} \space{\,}3\\ \times\phantom{0}4\space{\,}0 \\ \hline \phantom{\,}\end{array}$

Possible Answers:

1320

13200

1330

1310

Correct answer:

1320

Explanation:

In order to solve this problem using multiplication, we must multiply each digit of the multiplier by each digit of the multiplicand to get the answer, which is called the product.

$\frac{\begin{array}[b]{r}\textup{multiplicand}\\ \times\ \ \ \ \textup{multiplier}\end{array}}{\ \ \ \ \textup{product}}$

For this problem, 40 is the multiplier and 33 is the multiplicand. We start with the multiplier, and multiply the digit in the ones place by each digit of the multiplicand.

First, we multiply 0 and 3

$\begin{array}{r}\overset{~ \space{\ }}{\ 3} \space{\,}3\\ \times \phantom{0} 4\space{\,}0 \\ \hline \phantom{\,} \,0\end{array}$

Then, we multiply 0 and 3

$\begin{array}{r}\overset{~ \space{\ }}{\ 3} \space{\,}3\\ \times \phantom{0} 4\space{\,}0 \\ \hline \phantom{\,} \,0\,0\end{array}$

$\begin{array}{r}\overset{~ \space{\ }}{\ 3} \space{\,}3\\ \times \phantom{0} 4\space{\,}0 \\ \hline \phantom{\,} \,0\,0 \\ \, 0 \end{array}$

Next, we multiply 4 and 3

$\begin{array}{r}\overset{1 \space{\ }}{\ 3} \space{\,}3\\ \times \phantom{0} 4\space{\,}0 \\ \hline \phantom{\,} \,0\,0 \\ \, 2 \, 0\end{array}$

Then, we multiply 4 and 3and add the 1 that was carried

$\begin{array}{r}\overset{1 \space{\ }}{\ 3} \space{\,}3\\ \times \phantom{0} 4\space{\,}0 \\ \hline \phantom{\,} \,0\,0 \\ \, 1 \, 3 \, 2 \, 0\end{array}$

Finally, we add the two products together to find our final answer

$\begin{array}{r}\overset{1 \space{\ }}{\ 3} \space{\,}3\\ \times \phantom{0} 4\space{\,}0 \\ \hline \phantom{\,} \,0\,0 \\+\, 1 \, 3 \, 2 \, 0\\ \hline 1\, 3 \, 2\, 0\end{array}$

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Example Question #31 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Solve:

$\begin{array}{r}\overset{~ \space{\ }}{\ 4} \space{\,}1\\ \times\phantom{0}8\space{\,}3 \\ \hline \phantom{\,}\end{array}$

Possible Answers:

3393

34030

3413

3403

Correct answer:

3403

Explanation:

In order to solve this problem using multiplication, we must multiply each digit of the multiplier by each digit of the multiplicand to get the answer, which is called the product.

$\frac{\begin{array}[b]{r}\textup{multiplicand}\\ \times\ \ \ \ \textup{multiplier}\end{array}}{\ \ \ \ \textup{product}}$

For this problem, 83 is the multiplier and 41 is the multiplicand. We start with the multiplier, and multiply the digit in the ones place by each digit of the multiplicand.

First, we multiply 3 and 1

$\begin{array}{r}\overset{~ \space{\ }}{\ 4} \space{\,}1\\ \times \phantom{0} 8\space{\,}3 \\ \hline \phantom{\,} \,3\end{array}$

Then, we multiply 3 and 4

$\begin{array}{r}\overset{~ \space{\ }}{\ 4} \space{\,}1\\ \times \phantom{0} 8\space{\,}3 \\ \hline \phantom{\,}1\,2\,3\end{array}$

$\begin{array}{r}\overset{~ \space{\ }}{\ 4} \space{\,}1\\ \times \phantom{0} 8\space{\,}3 \\ \hline \phantom{\,}1\,2\,3 \\ \, 0 \end{array}$

Next, we multiply 8 and 1

$\begin{array}{r}\overset{~ \space{\ }}{\ 4} \space{\,}1\\ \times \phantom{0} 8\space{\,}3 \\ \hline \phantom{\,}1\,2\,3 \\ \, 8 \, 0\end{array}$

Then, we multiply 8 and 4

$\begin{array}{r}\overset{~ \space{\ }}{\ 4} \space{\,}1\\ \times \phantom{0} 8\space{\,}3 \\ \hline \phantom{\,}1\,2\,3 \\ \, 3\, 2\, 8 \, 0\end{array}$

Finally, we add the two products together to find our final answer

$\begin{array}{r}\overset{~ \space{\ }}{\ 4} \space{\,}1\\ \times \phantom{0} 8\space{\,}3 \\ \hline \phantom{\,}1\,2\,3 \\+\, 3\, 2\, 8 \, 0\\ \hline 3\, 4 \, 0\, 3\end{array}$

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Example Question #32 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Solve:

$\begin{array}{r}\overset{~ \space{\ }}{\ 2} \space{\,}7\\ \times\phantom{0}5\space{\,}5 \\ \hline \phantom{\,}\end{array}$

Possible Answers:

1485

1475

14850

1495

Correct answer:

1485

Explanation:

In order to solve this problem using multiplication, we must multiply each digit of the multiplier by each digit of the multiplicand to get the answer, which is called the product.

$\frac{\begin{array}[b]{r}\textup{multiplicand}\\ \times\ \ \ \ \textup{multiplier}\end{array}}{\ \ \ \ \textup{product}}$

For this problem, 55 is the multiplier and 27 is the multiplicand. We start with the multiplier, and multiply the digit in the ones place by each digit of the multiplicand.

First, we multiply 5 and 7

$\begin{array}{r}\overset{3 \space{\ }}{\ 2} \space{\,}7\\ \times \phantom{0} 5\space{\,}5\\ \hline \phantom{\,} \,5\end{array}$

Then, we multiply 5 and 2 and add the 3 that was carried

$\begin{array}{r}\overset{3 \space{\ }}{\ 2} \space{\,}7\\ \times \phantom{0} 5\space{\,}5 \\ \hline \phantom{\,}1\,3\,5\end{array}$

$\begin{array}{r}\overset{~ \space{\ }}{\ 2} \space{\,}7\\ \times \phantom{0} 5\space{\,}5 \\ \hline \phantom{\,}1\,3\,5 \\ \, 0 \end{array}$

Next, we multiply 5 and 7

$\begin{array}{r}\overset{3 \space{\ }}{\ 2} \space{\,}7\\ \times \phantom{0} 5\space{\,}5 \\ \hline \phantom{\,}1\,3\,5 \\ \, 5 \, 0\end{array}$

Then, we multiply 5 and 2and add the 3 that was carried

$\begin{array}{r}\overset{3 \space{\ }}{\ 2} \space{\,}7\\ \times \phantom{0} 5\space{\,}5 \\ \hline \phantom{\,}1\,3\,5 \\ \, 1 \, 3 \, 5 \, 0\end{array}$

Finally, we add the two products together to find our final answer

$\begin{array}{r}\overset{3 \space{\ }}{\ 2} \space{\,}7\\ \times \phantom{0} 5\space{\,}5 \\ \hline \phantom{\,}1\,3\,5 \\+\, 1 \, 3 \, 5 \, 0\\ \hline 1\, 4 \, 8\, 5\end{array}$

Report an Error

Example Question #555 : Number & Operations In Base Ten

Solve:

$\begin{array}{r}\overset{~ \space{\ }}{\ 5} \space{\,}2\\ \times\phantom{0}8\space{\,}2 \\ \hline \phantom{\,}\end{array}$

Possible Answers:

4254

4274

4264

42640

Correct answer:

4264

Explanation:

In order to solve this problem using multiplication, we must multiply each digit of the multiplier by each digit of the multiplicand to get the answer, which is called the product.

$\frac{\begin{array}[b]{r}\textup{multiplicand}\\ \times\ \ \ \ \textup{multiplier}\end{array}}{\ \ \ \ \textup{product}}$

For this problem, 82 is the multiplier and 52 is the multiplicand. We start with the multiplier, and multiply the digit in the ones place by each digit of the multiplicand.

First, we multiply 2 and 2

$\begin{array}{r}\overset{~ \space{\ }}{\ 5} \space{\,}2\\ \times \phantom{0} 8\space{\,}2 \\ \hline \phantom{\,} \,4\end{array}$

Then, we multiply 2 and 5

$\begin{array}{r}\overset{~ \space{\ }}{\ 5} \space{\,}2\\ \times \phantom{0} 8\space{\,}2 \\ \hline \phantom{\,}1\,0\,4\end{array}$

$\begin{array}{r}\overset{~ \space{\ }}{\ 5} \space{\,}2\\ \times \phantom{0} 8\space{\,}2 \\ \hline \phantom{\,}1\,0\,4 \\ \, 0 \end{array}$

Next, we multiply 8 and 2

$\begin{array}{r}\overset{1 \space{\ }}{\ 5} \space{\,}2\\ \times \phantom{0} 8\space{\,}2 \\ \hline \phantom{\,}1\,0\,4 \\ \, 6 \, 0\end{array}$

Then, we multiply 8 and 5and add the 1 that was carried

$\begin{array}{r}\overset{1 \space{\ }}{\ 5} \space{\,}2\\ \times \phantom{0} 8\space{\,}2 \\ \hline \phantom{\,}1\,0\,4 \\ \, 4 \, 1 \, 6 \, 0\end{array}$

Finally, we add the two products together to find our final answer

$\begin{array}{r}\overset{1 \space{\ }}{\ 5} \space{\,}2\\ \times \phantom{0} 8\space{\,}2 \\ \hline \phantom{\,}1\,0\,4 \\+\, 4 \, 1 \, 6 \, 0\\ \hline 4\, 2 \, 6\, 4\end{array}$

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All ISEE Middle Level Quantitative Resources

6 Diagnostic Tests 110 Practice Tests Question of the Day Flashcards Learn by Concept

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ISEE Middle Level Quantitative : ISEE Middle Level (grades 7-8) Quantitative Reasoning

Study concepts, example questions & explanations for ISEE Middle Level Quantitative

All ISEE Middle Level Quantitative Resources

Example Questions

Example Question #24 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Example Question #25 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Example Question #26 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Example Question #27 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Example Question #28 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Example Question #29 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Example Question #552 : Number & Operations In Base Ten

Example Question #31 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Example Question #32 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Example Question #555 : Number & Operations In Base Ten

All ISEE Middle Level Quantitative Resources

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